Isolated zeros on closure of a domain

Let $f$ be an analytic function on the open unit disk domain $D$. Suppose also that $f$ is bounded.

Since $f$ is bounded I believe that $f$ can be continuously extended to the closed unit disk. I know that the zeros of $f$ in the open disk $D$ are isolated. Are the zeros of $f$ in the closed unit disk also necessarily isolated?

Edit (March 5, 2011): My original answer gave as justification for claims about zero sets a result of Rudin and of Carleson that not only came a half a century after Fatou's result now mentioned below, but also didn't directly address the problem. I've edited to correct the deficiencies.

It is not true in general that $f$ can be extended continuously to the closed disk. The set of bounded analytic functions on the open disk is called the Hardy algebra, denoted $H^\infty$, and the set of analytic functions on the disk that have continuous extensions to the closed disk is called the disk algebra, often denoted $A(\mathbb{D})$, and $A(\mathbb D)$ is properly contained in $H^\infty$. Elements of the Hardy algebra do have radial limits (or even nontangential limits) a.e. on the boundary, allowing the Hardy algebra to be embedded in $L^\infty$ of the circle as the subspace of functions whose negatively indexed Fourier coefficients vanish.

Elements of the Hardy algebra must be nonzero a.e. on the circle (unless they are identically zero). In fact, this is even true for elements of $H^1$, i.e., for integrable functions on the circle whose negatively indexed Fourier coefficients vanish. (See Theorem 17.17 of Rudin's Real and complex analysis 3rd Edition for a much stronger statement.) So for elements of $A(\mathbb D)$, the zero sets on the boundary must be closed sets of Lebesgue measure zero. Fatou proved in 1906 that for every closed subset $E$ of the boundary with Lebesgue measure zero, there is an element of $A(\mathbb D)$ whose zero set is exactly $E$. In particular, this shows that the zeros need not be isolated. For example, you could consider Cantor-like subsets of the circle to obtain zero sets with very many limit points.

A related theorem of Rudin says that given any closed subset $E$ of the unit circle with (one-dimensional) Lebesgue measure 0, and given any complex valued continuous function $g$ on $E$, there is an element of the disk algebra whose restriction to $E$ is $g$. Here's the original 1956 article of Rudin on JSTOR. Carleson independently proved the theorem; here is a link (not open access) to his paper, published the following year. Rudin points out in his book Functional analysis on page 124 of the second edition that the result can be obtained as a corollary of a more general theorem of Bishop (Theorem 5.9 in the book) and the F. and M. Riesz theorem. Bishop's result is from the 1962 paper "A general Rudin-Carleson theorem."

A good reference for everything mentioned above is Hoffman's Banach spaces of analytic functions. In particular, Fatou's and Rudin's theorems appear in Chapter 6, followed by a characterization of the closed ideals of $A(\mathbb D)$.

For some explicit examples of bounded analytic functions on the open disk with no continuous extension to the closed disk (in other words, elements of $H^\infty\setminus A(\mathbb{D})$), consider Blaschke products with infinitely many zeros. If $f$ is such a function, then $f$ has boundary values of modulus 1 a.e., and since $f$ has infinitely many zeros, the set of zeros has a limit point $x$ on the circle. If $f$ had a continuous extension to the closed disk, this would imply respectively that $f$ has modulus 1 everywhere on the circle, while at the same time $f(x)=0$, a contradiction.

Jonas Meyer's answer goes much deeper, but let me say the following as well.

Even if your (nonzero -- you didn't say that, but of course you meant it!) function does extend continuously to the closed unit disk it need not have isolated zeros. Indeed, suppose that $f$ has infinitely many zeros in the open unit disk. Then the zero set of the (putative) extension of $f$ to the closed unit disk is an infinite subset of a compact space, so must have an accumulation point: i.e., there will be at least one point on the boundary which is a nonisolated zero.

In fact, by the Weierstrass Factorization Theorem, the zero set of an analytic function on the open unit disk can be any discrete subset without accumulation points in the open disk. You can choose such a set to have the entire boundary $|z| = 1$ of the disk contained in its closure. We conclude that there exists a nonzero analytic function $f$ on the open unit disk such that -- if it admits a continuous extension to the closed unit disk -- this extension is identically zero on the boundary.

I would be interested to know if this construction can be made unconditional, i.e., whether any analytic function with such a zero set can be extended continuously to the boundary.